How would you balance the following equation: I2+HNO3-->HIO3+NO2+H2O?

1 Answer
Nov 13, 2015

Answer:

This is a redox reaction, which may be approached by oxidation and reduction half equations.

Explanation:

Oxidation:
Elemental iodine is oxidized to #IO_3^-#, iodate.

#1/2I_2 +3H_2O rarr IO_3^(-) +5e^(-) + 6H^+# #(i)#

Reduction:

Nitric acid is reduced to nitric oxide #N^(V)rarrN^(IV)#:

#HNO_3 +H^+ +e^(-)rarr NO_2 +H_2O# #(ii)#

The overall redox reaction is #(i) + 5xx(ii)# #=# the overall redox equation:

#1/2I_2 + 5HNO_3 rarr IO_3^(-) +5NO_2 + 2H_2O +H^+#

Alternatively, I could write:

#1/2I_2 + 5HNO_3 rarr HIO_3 +5NO_2 + 2H_2O#

Are the final equations balanced with respect to mass and to charge? Don't trust my arithmetic. How would I remove the half coefficient on #I_2#?